Problem: Divide the following complex numbers. $ \dfrac{20-10i}{-3+4i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3-4i}$ $ \dfrac{20-10i}{-3+4i} = \dfrac{20-10i}{-3+4i} \cdot \dfrac{{-3-4i}}{{-3-4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(20-10i) \cdot (-3-4i)} {(-3+4i) \cdot (-3-4i)} = \dfrac{(20-10i) \cdot (-3-4i)} {(-3)^2 - (4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(20-10i) \cdot (-3-4i)} {(-3)^2 - (4i)^2} = $ $ \dfrac{(20-10i) \cdot (-3-4i)} {9 + 16} = $ $ \dfrac{(20-10i) \cdot (-3-4i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({20-10i}) \cdot ({-3-4i})} {25} = $ $ \dfrac{{20} \cdot {(-3)} + {-10} \cdot {(-3) i} + {20} \cdot {-4 i} + {-10} \cdot {-4 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{-60 + 30i - 80i + 40 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{-60 + 30i - 80i - 40} {25} = \dfrac{-100 - 50i} {25} = -4-2i $